将块矩阵的每个块与NumPy中的不同系数相乘

Question

假设我有一个(4*n)*(4*n)块矩阵，我想将(n*n)的每个块与相应的(4*4)矩阵的不同系数相乘-在NumPy中如何做到这一点?

例如:

>>> mat
matrix([[1, 1, 1, 1],
        [1, 1, 1, 1],
        [1, 1, 1, 1],
        [1, 1, 1, 1]])
>>> x
array([[1, 2],
       [3, 4]])

请求的结果应该是这样的:

>>> result
array([[1, 1, 2, 2],
       [1, 1, 2, 2],
       [3, 3, 4, 4],
       [3, 3, 4, 4]])

Answer 1

一种方法是从x:

构造(4*n)*(4*n)块矩阵:

result = np.multiply(mat, np.kron(x, np.ones((n,n))))

Answer 2

这是使用NumPy broadcasting -

的矢量化方法

m1,n1 = x.shape
m2,n2 = mat.shape
out = (mat.reshape(m1,m2//m1,n1,n2//n1)*x[:,None,:,None]).reshape(m2,n2)

示例运行-

In [41]: mat
Out[41]: 
array([[8, 8, 7, 2, 3, 4],
       [2, 4, 7, 5, 4, 8],
       [7, 2, 4, 5, 6, 5],
       [4, 3, 3, 6, 5, 3],
       [7, 3, 5, 8, 7, 7],
       [2, 5, 2, 4, 2, 7]])
In [42]: x
Out[42]: 
array([[1, 2],
       [3, 4]])
In [43]: out
Out[43]: 
array([[ 8,  8,  7,  4,  6,  8],
       [ 2,  4,  7, 10,  8, 16],
       [ 7,  2,  4, 10, 12, 10],
       [12,  9,  9, 24, 20, 12],
       [21,  9, 15, 32, 28, 28],
       [ 6, 15,  6, 16,  8, 28]])

我认为@Tim Fuchs的帖子中建议的np.kron将是这里最安全的选择，而不需要摆弄输入数组的大小。为了提高性能，下面是比较kron和broadcasting-based方法在适当大小的输入数组上的一些计时——

In [56]: mat = np.random.randint(2,9,(100,100))
In [57]: x = np.random.randint(2,9,(50,50))
In [58]: n = 2
In [59]: m1,n1 = x.shape # Ignoring timings from these as negligible
    ...: m2,n2 = mat.shape
    ...: 
In [60]: %timeit np.multiply(mat, np.kron(x, np.ones((n,n))))
1000 loops, best of 3: 312 µs per loop
In [61]: %timeit (mat.reshape(m1,m2//m1,n1,n2//n1)*x[:,None,:,None]).reshape(m2,n2)
10000 loops, best of 3: 83.7 µs per loop